Asked by Nyebuchi

A ship leaves a port and travels 21km on a bearing of 32 and then 45km on a bearing of 287 calculate its distance from the port and the bearing of the port from the ship
Answered by oobleck
The ship made a turn of 75 degrees when it changed course.
So, using the law of cosines, its distance d from port can be found via
d^2 = 21^2 + 45^2 - 2*21*45*cos75°
d = 44.5 km

The ship's final position is
28.97km W and 38.24km N of port
Now just turn that into a bearing - watch the direction you use -- from ship to port
Answered by henry2,
All angles are measured CW from +y-axis.
d = 21km[32o] + 45km[287o].

X = 21*sin32 + 45*sin287 = -31.91 km.
Y = 21*Cos32 + 45*Cos287 = 30.97 km.

d = -31.91 + 30.97i = 44.5km[-45.9o] = 44.5km[45.9o] W. of N. = 44.5km[314o] CW.

Tan A = X/Y.
Answered by Moshood oluwatobi
Sin 105 is cos -75
Answered by Kofi
That's good
Answered by olamilekan
give the diagram
Answered by Mary
Please I Need The Diagram
Answered by Ify
Thanks but I need the diagram



OK
Answered by Levi
Give working guide
Answered by Margret
I need the diagram of the above question